If $12$ persons are arranged in a row such that neither of two particular persons can sit on either end of the row, is
My attempt:
Total ways $=$ Sitting $12$ persons in a row $-$ Sitting $2$ particular persons $2$ ends of the row
$$=12!-2!\cdot 10!$$
But it seems that this answer is wrong.
Can anyone please explain to me the right answer. Thanks.
Best Answer
You have only removed from your count where both of your people sit at the ends and have neglected to remove where only one of your people sit at the ends but not the other.
Rather than thinking in terms of the perspective of the seats... we can think with respect to the people instead and approach with rule of product like usual.
Pick which seat the younger of the two special people sit (noting it cannot be either end): 10 choices
Pick which seat the older of the two special people sit (noting it cannot be either end or where the first person sat): 9 choices
Pick which seat the youngest of the remaining people sit (noting it cannot be where either of the first two people sat but now may include the ends): 10 choices
Pick which seat the next youngest of the remaining people sit (similar restrictions): 9 choices
Continue in this fashion seating each of the rest of the people in sequence
You will find then that there are $10\cdot 9\cdot 10!$ total arrangements.